Application of Derivatives 2 Question 29
####29. Let $f(x)=\begin{array}{cc}x e^{a x}, & x \leq 0 \\ x+a x^{2}-x^{3}, & x>0\end{array}$
where, $a$ is a positive constant. Find the interval in which $f^{\prime}(x)$ is increasing.
(1996, 3M)
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Answer:
Correct Answer: 29. (c, d)
Solution:
- NOTE This type is asked in 1983 and repeat after 13 years.
At $x=0$, LHL $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x e^{a x}=0$ and $\quad$ RHL $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x+a x^{2}-x^{3}\right)=0$
Therefore, $\mathrm{LHL}=\mathrm{RHL}=0=f(0)$
So, $f(x)$ is continuous at $x=0$.
Also, $\quad f^{\prime}(x)=\begin{aligned} & 1 \cdot e^{a x}+a x e^{a x}, \text { if } \quad x<0 \\ & 1+2 a x-3 x^{2}, \text { if } \quad x>0\end{aligned}$
and $\quad L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$ =\lim _{x \rightarrow 0^{-}} \frac{x e^{a x}-0}{x}=\lim _{x \rightarrow 0^{-}} e^{a x}=e^{0}=1 $
and
$ \begin{aligned} R f^{\prime}(0) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x+0} \\ & =\lim _{x \rightarrow 0^{+}} \frac{x+a x^{2}-x^{3}-0}{x} \\ & =\lim _{x \rightarrow 0^{+}} 1+a x-x^{2}=1 \end{aligned} $
Therefore, $L f^{\prime}(0)=R f^{\prime}(0)=1 \quad \Rightarrow \quad f^{\prime}(0)=1$ Hence, $\quad f^{\prime}(x)=\begin{array}{cc}(a x+1) e^{a x}, & \text { if } x<0 \\ 1, & \text { if } x=0 \\ 1+2 a x-3 x^{2}, & \text { if } x>0\end{array}$
Now, we can say without solving that, $f^{\prime}(x)$ is continuous at $x=0$ and hence on $R$. We have,
and
$ f^{\prime \prime}(x)=\begin{array}{ll} a e^{a x}+a(a x+1) e^{a x}, & \text { if } x<0 \\ 2 a-6 x, & \text { if } x>0 \end{array} $
$ L f^{\prime \prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x-0} $
$ =\lim _{x \rightarrow 0^{-}} \frac{(a x+1) e^{a x}-1}{x} $
$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} a e^{a x}+\frac{e^{a x}-1}{x} \\ & =\lim _{x \rightarrow 0^{-}} a e^{a x}+a \cdot \lim _{x \rightarrow 0^{-}} \frac{e^{a x}-1}{a x} \\ & =a e^{0}+a(1)=2 a \end{aligned} $
and
$ \begin{aligned} R f^{\prime \prime}(0) & =\lim _{x \rightarrow 0^{+}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x+0} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\left(1+2 a x-3 x^{2}\right)-1}{x} \\ & =\lim _{x \rightarrow 0^{+}} \frac{2 a x-3 x^{2}}{x}=\lim _{x \rightarrow 0^{+}} 2 a-3 x=2 a \end{aligned} $
Therefore, $L f^{\prime \prime}(0)=R f^{\prime \prime}(0)=2 a$
Hence, $\quad f^{\prime \prime}(x)=\begin{array}{ll}a(a x+2) e^{a x}, & \text { if } x<0 \\ 2 a, & \text { if } x=0 \\ 2 a-6 x, & \text { if } x>0\end{array}$
Now, for $x<0, f^{\prime \prime}(x)>0$, if $a x+2>0$
$\Rightarrow$ For $x<0, f^{\prime \prime}(x)>0$, if $x>-2 / a$
$\Rightarrow \quad f^{\prime}(x)>0$, if $-\frac{2}{a}<x<0$
and for $x>0, f^{\prime \prime}(x)>0, \quad$ if $2 a-6 x>0$
$\Rightarrow$ for $x>0, \quad f^{\prime \prime}(x)>0$, if $x<a / 3$
Thus, $f(x)$ increases on $[-2 / a, 0]$ and on $[0, a / 3]$.
Hence, $f(x)$ increases on $-\frac{2}{a}, \frac{a}{3}$.
